f(g(

p(0) -> g(0)

g(s(p(

R

↳Dependency Pair Analysis

F(g(x), g(y)) -> F(p(f(g(x), s(y))), g(s(p(x))))

F(g(x), g(y)) -> P(f(g(x), s(y)))

F(g(x), g(y)) -> F(g(x), s(y))

F(g(x), g(y)) -> G(s(p(x)))

F(g(x), g(y)) -> P(x)

P(0) -> G(0)

Furthermore,

R

↳DPs

→DP Problem 1

↳Remaining Obligation(s)

The following remains to be proven:

**F(g( x), g(y)) -> F(p(f(g(x), s(y))), g(s(p(x))))**

f(g(x), g(y)) -> f(p(f(g(x), s(y))), g(s(p(x))))

p(0) -> g(0)

g(s(p(x))) -> p(x)

Duration:

0:01 minutes